# If (7x^2)/(2x-3), what are the points of inflection, concavity and critical points?

Apr 26, 2017

There is a local maximum at $\left(0 , 0\right)$ and a local minimum at $\left(3 , 21\right)$
No inflection point.
The function is concave in the interval $\left(- \infty , \frac{3}{2}\right)$ and convex in the interval $\left(\frac{3}{2} , + \infty\right)$

#### Explanation:

We need

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$

We calculate the first and second derivatives

Let $f \left(x\right) = \frac{7 {x}^{2}}{2 x - 3}$

$u = 7 {x}^{2}$, $\implies$, $u ' = 14 x$

$v = 2 x - 3$, $\implies$, $v ' = 2$

Therefore,

$f ' \left(x\right) = \frac{14 x \left(2 x - 3\right) - 7 {x}^{2} \left(2\right)}{2 x - 3} ^ 2$

$= \frac{28 {x}^{2} - 42 x - 14 {x}^{2}}{2 x - 3} ^ 2$

$= \frac{14 {x}^{2} - 42 x}{2 x - 3} ^ 2$

$= \frac{14 x \left(x - 3\right)}{2 x - 3} ^ 2$

The critical points are when $x = 0$ and $x = 3$

We can build a chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a a}$$\frac{3}{2}$$\textcolor{w h i t e}{a a a a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 3$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$↗$\textcolor{w h i t e}{a a a a}$↘$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$↘$\textcolor{w h i t e}{a a a a}$↗

There is a local maximum at $\left(0 , 0\right)$ and a local minimum at $\left(3 , 21\right)$

Now, we determine the second derivative

$u = \left(14 {x}^{2} - 42 x\right)$, $\implies$, $u ' = 28 x - 42$

$v = {\left(2 x - 3\right)}^{2}$, $\implies$, $v ' = 4 \left(2 x - 3\right)$

$f ' ' \left(x\right) = \frac{\left(28 x - 42\right) {\left(2 x - 3\right)}^{2} - 4 \left(14 {x}^{2} - 42 x\right) \left(2 x - 3\right)}{2 x - 3} ^ 4$

$= \frac{\left(2 x - 3\right) \left(56 {x}^{2} - 84 x - 84 x + 126 - 56 {x}^{2} + 168 x\right)}{2 x - 3} ^ 4$

$= \frac{126}{2 x - 3} ^ 3$

There is no inflexion point.

We can build a chart

$\textcolor{w h i t e}{a a a a}$$I n t e r v a l$$\textcolor{w h i t e}{a a a a}$$\left(- \infty , \frac{3}{2}\right)$$\textcolor{w h i t e}{a a a a}$$\left(\frac{3}{2} , + \infty\right)$

$\textcolor{w h i t e}{a a a a}$$f ' ' \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a a a a}$$\cap$$\textcolor{w h i t e}{a a a a a a a a a a a}$$\cup$

The function is concave in the interval $\left(- \infty , \frac{3}{2}\right)$ and convex in the interval $\left(\frac{3}{2} , + \infty\right)$

graph{7x^2/(2x-3) [-66.14, 65.5, -22.55, 43.3]}